Note that the sum of the two equations (1) x ^ 2-3x + 2 = 0 is A1 (2) x ^ 2 + 7x + 12 = 0, the sum of the two equations is A2 (3) x ^ 2-11x + 30 = 0, and the sum of the two equations is A3 to find a2006 Note that the sum of the two equations (1) x ^ 2-3x + 2 = 0 is A1 (2) X ^ 2 + 7x + 12 = 0, the sum of the two is A2 (3) The sum of x ^ 2-11x + 30 = 0 is A3 Find the value of a2006 Find a1 + A2 + a3 +a2008

Note that the sum of the two equations (1) x ^ 2-3x + 2 = 0 is A1 (2) x ^ 2 + 7x + 12 = 0, the sum of the two equations is A2 (3) x ^ 2-11x + 30 = 0, and the sum of the two equations is A3 to find a2006 Note that the sum of the two equations (1) x ^ 2-3x + 2 = 0 is A1 (2) X ^ 2 + 7x + 12 = 0, the sum of the two is A2 (3) The sum of x ^ 2-11x + 30 = 0 is A3 Find the value of a2006 Find a1 + A2 + a3 +a2008

According to Weida's theorem X1 + x2 = - B / a | A1 = 3, A2 = - 7, A3 = 11, there are | A2 | - | A1 | = | A3 | - | A2 | = 4 | - an | = 4n-1an = (4N-1) * (- 1) ^ (n + 1), n ∈ n *. A2006 = (4 * 2006-1) (- 1) ^ (2006 + 1) = - 8023s2008 = (a1 + A2) + (A3 + A4) + +(a2007+a2008)=(-4)*2008/2= -4016...

The system of equations {x-3y = the square of 3x - the square of 3Y {7x + y = the square of 10x - the square of 10Y
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X-3y = 3x & # 178; - 3Y & # 178; (1) 7x + y = 10x & # 178; - 10Y & # 178; (2) (1) multiply 10 to get 10x-30y = 30x & # 178; - 30y & # 178; (3) (2) multiply 3 to get 21x + 3Y = 30x & # 178; - 30y & # 178; (4) (3) (4) combine to get 10x-30y = 21x + 3Y, that is, substituting x = - 3Y into (1) to get - 3y-3y = 3 (- 3Y) &# 178; -

3 / x-0.3 / 1-3x = 7x and 4 / 3y-1 = 1-8 / 3-y and 0.4 / 3x-0.4 + 0.5 = 2 / 15 + 4x
How to do 1 / 2-A times - 2 = 2 / 3-2 + 2A

-The original formula of 2 (1 / 2-A) = - 2 / 3 + 2a is - 1 + 2A = - 2 / 3 + 2A